How to Design a Variable Regulated Power Supply

The electronics devices works in dc power supply comes with different voltage specifications. Some devices works in battery power with voltages ranging from 1.5V to 9V, but majority of the devices can be directly plugged into a power plug. Such devices have a built in AC to low voltage DC converter inside it and the voltage output of that converter will be corresponding to the design of the entire circuit. The design of a circuit actually begins with considering the value of the voltage which will be supplied to it.

The circuits which work in 5V power supply are most common which contains 5V TTL devices, but there is other category of TTL devices which works in 3.3V supply. Apart from these two categories there are CMOS based circuits and there is all kind of analog circuits which are designed to work on different voltages. Those who are interested in working with all these kind of circuits should have a variable power supply source. Since most of the electronic designs demands regulated power supply the variable power source should be regulated also. This article discusses how to design a Variable Regulated Power Supply.

AIM: design variable regulated power supply from 1V to 25 V @ 1A

We shall follow same procedure followed in Regulated Power Supply.

Design of Regulated section:-

Step 1: Selecting regulator chip:

Because here we need variable supply we shall use LM317. The equation for output voltage is

V_o = 1.25 (1 + R₂ / R₁)

In above equation R₁ is fixed as 240?. So by varying R₂ we can change output voltage. Because required o/p voltage is 25 V, we shall calculate required value of R₂as

R₂ = (V_o/1.25 – 1) × R₁

= (25 / 1.25 – 1) × 240

= 4560 ?

So we shall use 4.7 K ? potentiometer

Step 2: Selecting filter capacitor:

We may neglect input capacitor filter. For output we can use either 0.1 µF or 1 µF capacitor Output capacitor is required to suppress any spike or glitch in fixed output voltage that may occur due to transient change in AC input.

Design of unregulated section –

If we follow same calculations as given in previous design then we have following values

V_Lmin = 28 V

V_o = 5 V

V_dcFL = 30.5

V_dcNL = 38.5

From these we shall find out transformer, capacitor and diode ratings

Step 3: Selecting transformer:

E_rms= V_dcNL/ 1.41

= 38.5 / 1.41

= 27.30 VAC

So we may select Center tap transformer of 15 – 0 – 15. Current rating for secondary of transformer should be at least 1.8 I_L. That means the current rating can be 2 A.

Finally select transformer with

T = 230 / 30 VAC @ 2A

Step 4: Selecting diode

Selecting diode means finding current capacity and PIV of diode.

1. Current capacity I_C> I_Lthat means Ic can be 1 A or more

2. PIV = V_dcNL+ 20%V_dcNL= 46.2. again going for higher value that is 50 V

Finally required diodes are with

D = 1A @ 50V

Step 5: Selecting capacitor

Assumed value of capacitor is C = 1000 µF.

WLDC = V_dcNL+ 20%V_dcNL= 46.2. again going for higher value that is 50 V

Finally required capacitor is with